The lattice enthalpy of an ionic compound is the enthalpy change when one mole of an ionic compound in its solid state dissociates into its ions in the gaseous state. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure the lattice enthalpy of $NaCl_{(s)}$.

(A) The lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in the gaseous state. For the reaction:
$NaCl_{(s)} \rightarrow Na_{(g)}^{+} + Cl_{(g)}^{-}$; $\Delta_{lattice} H^{\ominus} = +788 \ kJ \ mol^{-1}$
Since it is impossible to determine lattice enthalpies directly by experiment,we use an indirect method by constructing an enthalpy diagram known as the $Born-Haber$ cycle.
We calculate the lattice enthalpy of $NaCl_{(s)}$ using the following steps:
$(I)$ $Na_{(s)} \rightarrow Na_{(g)}$; Sublimation of sodium metal,$\Delta_{sub} H^{\ominus} = 108.4 \ kJ \ mol^{-1}$
$(II)$ $Na_{(g)} \rightarrow Na_{(g)}^{+} + e_{(g)}^{-}$; Ionization of sodium atoms,$\Delta_{i} H^{\ominus} = 496 \ kJ \ mol^{-1}$
$(III)$ $\frac{1}{2} Cl_{2(g)} \rightarrow Cl_{(g)}$; Dissociation of chlorine,$\frac{1}{2} \Delta_{bond} H^{\ominus} = 121 \ kJ \ mol^{-1}$
$(IV)$ $Cl_{(g)} + e_{(g)}^{-} \rightarrow Cl_{(g)}^{-}$; Electron gain by chlorine atoms,$\Delta_{eg} H^{\ominus} = -348.6 \ kJ \ mol^{-1}$
According to Hess's Law,the total enthalpy change for the formation of $NaCl_{(s)}$ from its elements is the sum of these individual steps,allowing us to solve for the lattice enthalpy.